Question #fd6e0

1 Answer
Oct 3, 2016

Answer:

Approx. #30*mg# #O_2# gas are required.

Explanation:

We need a stoichiometrically balanced equation:

#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)#

Is this balanced? Why?

The equation tells us unequivocally that 1 mole of methane gas requires 2 moles of dioxygen gas for complete combustion.

#"Moles of methane"# #=# #(7.90xx10^-3*g)/(16.04*g*mol^-1)# #=4.93xx10^-4*mol#.

And thus 2 equiv of dioxygen gas are required (I use the term #"dioxygen"# to avoid confusion with regard to the number of oxygen atoms I need) per equiv of methane gas.

#"Moles of dioxygen"# #=# #4.93xx10^-4*cancel(mol)xx2xx32.00*g*cancel(mol^-1)=??g#